 # NCERT Solutions for Square roots Class 8 Maths Chapter 6 CBSE Part 2

In this page we have NCERT Solutions for Square roots Class 8 Maths Chapter 6 for EXERCISE 2 . Hope you like them and do not forget to like , social share and comment at the end of the page.
Question 1
Find the square of the following numbers.
(i) 32
(ii) 35
(iii) 86
(iv) 93
(v) 71
(vi) 46
1) 322
We can find the square using direct multiplication
= 32 x 32 = 1024
But above method can be cumbersome to calculate. We  can calculate such values in the another better way f
Since, 32 can be written as (30+2)
So, 322 = (30+2)2 = (30+2)(30+2)
Now we know the identity
(a+b)2= a2 + b2 +2ab
= 302 + 2 x 30 x 2 + 22
= 900 + 120  + 4 = 1024
2) (35)2 = (30+5)2
Solving on similar lines as above
= 302 + 2 x 30 x 5  + 52
= 900 + 300 + 25 = 1225
3) 862 = (80 + 6)2
= 802 + 2 x  80 x 6  + 62
= 6400 + 960 + 36 = 7396
4) 932 = (90+3)2
= 90 2 +2 x 90 x 3  + 3 2
= 8100 + 540 + 9 = 8649
5) 71 2 = (70 + 1) 2
= 702 +2 x 70 x 1  + 1 x 1
= 4900 + 140 + 1 = 5040 + 1 = 5041
6) 462 = (40+6)2
= 40 2 + 2 x 40 x 6  + 62
= 1600 + 480 + 36 = 2080 + 36 = 2116

Question: 2
Write a Pythagorean triplet whose one member is:
(i) 6
(ii) 14
(iii) 16
(iv) 18
As we know 2n, n 2 + 1 and n2 - 1 form a Pythagorean triplet for any number, n > 1.
1) If we take n 2 + 1 or  n2 - 1 to be 6 then then the value of n will  not integer(n2  will be 5 or 7)
So  we can  2n = 6
Therefore, n = 3
And, n2 + 1 = 3 2 + 1= 9 + 1 = 10
And,n 2 - 1 = 3 2 - 1 = 9 - 1 = 8
Test: 6 2 + 8 2 = 36 + 64 = 100 = 102
Hence, the triplet is 6, 8, and 10 Answer
2) If we take n 2 + 1 or  n2 - 1 to be 14 then then the value of n will  not integer(n2  will be 15 or 13)
So we can take 2n= 14, therefore, n = 7
Now, n 2 + 1 = 7 2 + 1 = 49 + 1 = 50
And, n2 - 1 = 7 2 - 1 = 49 - 1 = 48
Test: 14 2 + 48 2 = 196 + 1304 = 2500 = 50 2
Hence, the triplet is 14, 48, and 50 Answer
3) If we take n 2 + 1 or  n2 - 1 to be 16 then then the value of n will  not integer(n2  will be 17 or 15)
Let us assume 2n = 16, then n = 8
Now, n2 + 1 = 8 2 + 1 = 64 + 1 = 65
And, n 2 - 1 = 8 2 - 1 = 64 - 1 = 63
Test: 162 + 63 2 = 256 + 3969 = 4225 = 65 2
Hence, the triplet is 16, 63, and 65 Answer
4) If we take n 2 + 1 or  n2 - 1 to be 18 then then the value of n will  not integer(n2  will be 19 or 17)
Let us assume 2n = 18, therefore, n = 9
Now, n 2 + 1 = 9 2 + 1 = 81 + 1 = 82
And, n 2 - 1 = 9 2 - 1 = 81 - 1 = 80
Test: 18 2 + 80 2 = 324 + 6400 = 6724 = 82 2

• Notes
• Worksheets & Ncert Solutions

Reference Books for class 8 Math

Given below are the links of some of the reference books for class 8 Math.

1. Mathematics Foundation Course for JEE/Olympiad : Class 8 This book can take students maths skills further. Only buy if child is interested in Olympiad/JEE foundation courses.
2. Mathematics for Class 8 by R S Aggarwal Detailed Mathematics book to clear basics and concepts. I would say it is a must have book for class 8 student.
3. Pearson Foundation Series (IIT -JEE / NEET) Physics, Chemistry, Maths & Biology for Class 8 (Main Books) | PCMB Combo : These set of books could help your child if he aims to get extra knowledge of science and maths. These would be helpful if child wants to prepare for competitive exams like JEE/NEET. Only buy if you can provide help to the child while studying.
4. Reasoning Olympiad Workbook - Class 8 :- Reasoning helps sharpen the mind of child. I would recommend students practicing reasoning even though they are not appearing for Olympiad.

You can use above books for extra knowledge and practicing different questions.

Note to our visitors :-

Thanks for visiting our website.